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Evaluate! surface integral over surface

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the surface integral of G over the surface S
    S is the parabolic cylinder y=2x^2, 0=< x =<5, 0=< z =<5
    G(x,y,z)=6x

    Answer is one of the following:
    1. (15/8)*(401sqrt(401)-1)
    2. (5/8)*(401sqrt(401)-1)
    3. (15/8)*(401sqrt(401)+1)
    4. (5/8)*(401sqrt(401)+1)

    2. Relevant equations


    3. The attempt at a solution
    Let f(x,y,z)=y-2x^2=0
    n=gradf/||gradf||=(-4xi+1j+0k)/sqrt(16x^2+1)
    n*=j
    G.n=-24x^2/sqrt(16x^2+1)
    n.n*=1/sqrt(16x^2+1)
    therefore the double integral over S = SS (G.n/n.n*) dzdx
    solving the double integral gets -5000
     
  2. jcsd
  3. May 29, 2012 #2
    Let me try ot reason through what you have. If at the end you were integrating dzdx, I'll assume I should try to paramet(e)rize the surface in x and z.

    Then r(x,z)=(x,2x^2,z). Then r_x=(1,4x,0), while r_z=(0,0,1).

    Then cross product is (r_x)x(r_z)=(4x,-1,0).

    Then the infinitesimal area on the surface is given by

    dS=sqrt(16x^2+1)dxdz.

    Now we want to integrate the SCALAR G(x,y,z)=4x against the area.

    That is, integrate G dS.

    So [itex]\int_{x=0}^5\int_{z=0}^54x\sqrt{16x^2+1}\ dzdx.[/itex]

    So my first guess is, you are mixing up VECTOR integrals with SCALAR integrals. You might compare them and their derivations.
     
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